Bezout's Theorem for weighted homogeneous polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:25:17Z http://mathoverflow.net/feeds/question/80933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80933/bezouts-theorem-for-weighted-homogeneous-polynomials Bezout's Theorem for weighted homogeneous polynomials Stanley Yao Xiao 2011-11-14T21:27:08Z 2011-11-15T13:48:48Z <p>Bezout's Theorem states that for two homogeneous polynomials $f(x,y,z), g(x,y,z)$ over an algebraically closed field of degrees $m,n$ respectively, such that the two polynomials do not share a common component, then the number of intersections of $f,g$ is equal to $mn$ counting multiplicity. Is there an analogue of this theorem for WEIGHTED homogeneous polynomials? That is, suppose that $w_1, w_2, w_3$ are three coprime positive integers, and for a given polynomial $h(x,y,z)$ let $e_1(h), e_2(h), e_3(h)$ denote the degrees of $x,y,z$ in $h$ respectively. We say that $h$ is weighted homogeneous of degree $d$ with weight $(w_1, w_2, w_3)$ if $h$ satisfies $w_1 e_1(h) + w_2 e_2(h) + w_3 e_3(h) = d$. If we allow $d$ to vary across all positive integers, then the resulting set of polynomials is the set of weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$. </p> <p>So my question is, is there an analogue to Bezout's Theorem in this setting? That is, is there a constant $W = W(w_1, w_2, w_3)$ which depends on $w_1, w_2, w_3$ such that if $f,g$ are two weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$ with no common componets, then the number of intersections of $f,g$ is bounded by $W(w_1, w_2, w_3) \deg(f) \deg(g)$?</p> http://mathoverflow.net/questions/80933/bezouts-theorem-for-weighted-homogeneous-polynomials/80935#80935 Answer by Francesco Polizzi for Bezout's Theorem for weighted homogeneous polynomials Francesco Polizzi 2011-11-14T22:09:04Z 2011-11-15T13:48:48Z <p>The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.</p> <p>In the case of weighted homogeneous polynomials one needs a Bézout theorem in the <em>weighted projective plane</em> $\mathbb{P}^2(w_1, w_2, w_3)$. Such a result can be found, for instance, in the paper by Bartolo, Martin-Morales and Ortigas-Galindo <a href="http://riemann.unizar.es/~jortigas/Pau-zgz.pdf" rel="nofollow">Q-resolutions and intersection numbers</a>, Section 5. It turns out that the intersection number of two curves of equation $f=0$ and $g=0$ in $\mathbb{P}^2(w_1, w_2, w_3)$ is given by <code>$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$</code> where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.</p> <p>Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a <em>singular</em> variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are <em>not</em> Cartier. Consequently, one can obtain a <em>rational</em> intersection number, insted of an integer one. </p> <p>For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$ is equal to $\frac{1 \cdot 1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$.</p> <p>This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex $[0:0:1]$) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Now two hyperplane sections intersect in two points, so we have $(2L_1)(2L_2)=2$, that is $L_1L_2=\frac{1}{2}$.</p>